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In the figure, $X$ is a point in the interior of square $A B C D$. $AXYZ$ is also a square. If $D Y=3\ cm$ and $AZ=2 \ cm$. Then $BY=?$."

Given: In the figure, $X$ is a point in the interior of square $A B C D$. $AXYZ$ is also a square. If $D Y=3\ cm$ and $AZ=2 \ cm$. 

To do: To find $BY=?$

Solution:


 

As given $ABCD$ is a square. And $AXYZ$ is also a squre.

$\Rightarrow AX=XY=YZ=AZ=2\ cm$ and $\angle A=\angle X=\angle Y=\angle Z=90^o$

And $DY=3\ cm$      [Given]

$\Rightarrow DZ=DY+DZ=3+2=5\ cm$

In $\vartriangle ADZ$,

$\because \angle=90^o \Rightarrow \vartriangle ADZ$ is right angled.

On using Pythagoras theorem 

$AD^2=DZ^2+AZ^2$

$\Rightarrow AD^2=5^2+2^2$

$\Rightarrow AD^2=25+4$

$\Rightarrow AD^2=29$

$\Rightarrow AD=\sqrt{29}\ cm$

$\Rightarrow AD=AB=BC=CD=\sqrt{29}\ cm$

$\vartriangle ABX$ is also right angled.

Using Pythagoras theorem:

$AB^2=AX^2+BX^2$

$\Rightarrow BX^2=AB^2-AX^2$

$\Rightarrow BX^2=( \sqrt{29})^2-2^2$

$\Rightarrow BX^2=29-4$

$\Rightarrow BX^2=25$

$\Rightarrow BX=\sqrt{25}$

$\Rightarrow BX=5\ cm$

Thus, $BY=BX+XY=5+2=7\ cm$.

Updated on: 10-Oct-2022

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